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Math Help - Arc Length Problem

  1. #16
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    Re: Arc Length Problem

    Because you didn't listen to what I was saying. What angle of $\displaystyle \begin{align*} \theta \end{align*}$ makes $\displaystyle \begin{align*} \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}$?
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  2. #17
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    Re: Arc Length Problem

    \arcsin(\dfrac{1}{\sqrt{2}}) = 45 = \dfrac{\pi}{4}

    45 degrees or \dfrac{\pi}{4}, but is that the answer to the whole problem?


    Quote Originally Posted by Prove It View Post
    Because you didn't listen to what I was saying. What angle of $\displaystyle \begin{align*} \theta \end{align*}$ makes $\displaystyle \begin{align*} \sin{ \left( \theta \right) } = \frac{1}{\sqrt{2}} \end{align*}$?
    Last edited by Jason76; July 20th 2014 at 10:36 AM.
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  3. #18
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    Re: Arc Length Problem

    You have been given very good hints and responses here and have NOT done what was suggested.

    Again, if \sqrt{2}sin(\theta)= x= 0, what is \theta? If \sqrt{2}sin(\theta)= x= 1 what is \theta?
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  4. #19
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    Re: Arc Length Problem

    \sqrt{2 - x^{2}}

    0\leq x \leq 1

    y = (2 - x^{2})^{2}

    y' = -x(2 - x^{2})^{-1/2}

    \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}

    \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x}}

    Let's stop here, and really look at the next move. Because I still don't really understand this part.
    Last edited by Jason76; July 20th 2014 at 10:11 PM.
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  5. #20
    Super Member Matt Westwood's Avatar
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    Re: Arc Length Problem

    Quote Originally Posted by Jason76 View Post
    \sqrt{2 - x^{2}}

    0\leq x \leq 1

    y = (2 - x^{2})^{2}

    y' = -x(2 - x^{2})^{-1/2}

    \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}}

    \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x}}

    Let's stop here, and really look at the next move. Because I still don't really understand this part.
    Put it over a common denominator:

    \int_{0}^{1} \sqrt{+ \dfrac{(2 - x) + x^2}{2 - x}} dx

    And don't forget your differential. It is vital that you do not forget your differential.
    Last edited by Matt Westwood; July 21st 2014 at 12:52 AM.
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  6. #21
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    Re: Arc Length Problem

    Find arc length on interval 0 \leq x \leq 1

    \sqrt{2 - x^{2}}

    0\leq x \leq 1

    y = (2 - x^{2})^{2}

    y' = -x(2 - x^{2})^{-1/2}

    L = \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}} dx

    L = \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x^{2}}}dx

    Get a common denominator. Eventually you get the next line (after you get the common denominator and add everything up)

    L = \int_{0}^{1} \sqrt{\dfrac{2}{2 - x^{2}}} dx

    Preparing for trig substitution, take the constant out.

    L = \sqrt{2} \int_{0}^{1} \sqrt{\dfrac{1}{2 - x^{2}}} dx

    Now time for trig substitution.

     a = 2

     x = 2\sin\theta

     dx = 2\cos\theta

    4 - (2\sin\theta)^{2} = 4 - 4\sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4\cos^{2}\theta

    tex]L = \sqrt{2} \int_{0}^{1} \dfrac{1(2\cos\theta)}{\sqrt{4\cos^{2}\theta}}[/tex]

    L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{\sqrt{4\cos^{2}\theta}}

    L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{2\cos \theta}

    L = \int_{0}^{1} \sqrt{2} What to do at this point because no trig intervals to work with? Change limits? We looked at this before, but still don't understand.
    Last edited by Jason76; July 22nd 2014 at 09:59 PM.
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  7. #22
    Super Member Matt Westwood's Avatar
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    Re: Arc Length Problem

    Quote Originally Posted by Jason76 View Post
    Find arc length on interval 0 \leq x \leq 1

    \sqrt{2 - x^{2}}

    0\leq x \leq 1

    y = (2 - x^{2})^{2}

    y' = -x(2 - x^{2})^{-1/2}

    L = \int_{0}^{1} \sqrt{1 + (x(2 - x^{2})^{-1/2})^{2}} dx

    L = \int_{0}^{1} \sqrt{1 + \dfrac{x^{2}}{2 - x^{2}}}dx

    Get a common denominator. Eventually you get the next line (after you get the common denominator and add everything up)

    L = \int_{0}^{1} \sqrt{\dfrac{2}{2 - x^{2}}} dx

    Preparing for trig substitution, take the constant out.

    L = \sqrt{2} \int_{0}^{1} \sqrt{\dfrac{1}{2 - x^{2}}} dx

    Now time for trig substitution.

     a = 2

     x = 2\sin\theta

     dx = 2\cos\theta

    4 - (2\sin\theta)^{2} = 4 - 4\sin^{2}\theta = 4(1 - \sin^{2}\theta) = 4\cos^{2}\theta

    tex]L = \sqrt{2} \int_{0}^{1} \dfrac{1(2\cos\theta)}{\sqrt{4\cos^{2}\theta}}[/tex]

    L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{\sqrt{4\cos^{2}\theta}}

    L = \sqrt{2} \int_{0}^{1} \dfrac{2\cos\theta}{2\cos \theta}

    L = \int_{0}^{1} \sqrt{2} What to do at this point because no trig intervals to work with? Change limits? We looked at this before, but still don't understand.
    First note that the first mistake you made was in the line under "Now time for trig substitition". You have:


    L = \sqrt 2 \int_0^1 \dfrac 1{\sqrt{2 - x^2} } dx

    In order to make the substitution you want, you need to see this in the form:

    L = \sqrt 2 \int_0^1 \dfrac 1{\sqrt{a^2 - x^2} } dx

    You have set a = 2 whereas instead you needed .... ?

    Oh, and be careful with your indices, you've written:
    y = (2 - x^{2})^{2}
    up the top there, which is wrong.
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